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Unformatted text preview: practice 09 Due: Mar 27 2007, 1:00 pm 1 Question 1, chap 11, sect 3. part 1 of 1 8 points A disc of mass m moves horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 7 m . The second disc is moving horizon tally to the right with speed v 2 at the moment of impact. m 7 m v v 2 before The two discs stick together upon impact. 8 m v f after What is the speed of the composite body immediately after the collision? 1. v f = 2 7 v 2. v f = 7 27 v 3. v f = 5 8 v 4. v f = 9 16 v correct 5. v f = 1 5 v 6. v f = 11 32 v 7. v f = 3 5 v 8. v f = 13 40 v 9. v f = 11 35 v 10. None of these are correct. Explanation: The total momentum of the system is con served because there is no exterior force. So what we have is ( m + 7 m ) v f = mv + 7 m v 2 8 mv f = mv parenleftbigg 1 + 7 2 parenrightbigg 8 v f = parenleftbigg 2 2 + 7 2 parenrightbigg v 8 v f = 9 2 v v f = 9 (2) (8) v = 9 16 v , so = 9 16 v . Question 2, chap 10, sect 99. part 1 of 3 8 points Two small spheres of mass 478 g and 611 g are suspended from the ceiling at the same point by massless strings of equal length 10 . 2 m. The lighter sphere is pulled aside through an angle of 49 from the vertical and let go. The acceleration of gravity is 9 . 8 m / s 2 . 9 . 8 m / s 2 Before 478 g 1 . 2 m 49 611 g practice 09 Due: Mar 27 2007, 1:00 pm 2 9 . 8 m / s 2 After f At what speed will the lighter mass m 1 hit the heavier mass m 2 ? Correct answer: 8 . 29221 m / s (tolerance 1 %). Explanation: Let : m 1 = 478 g , m 2 = 611 g , i = 49 , = 10 . 2 m , v i = before collision , V f = after collision , and g = 9 . 8 m / s 2 . The velocity just before the collision v i can be determined by energy conservation. When particle 1 is at its initial condition, it is at rest and displaced by an angle i from the vertical. The total energy is all potential and is given by U i = m 1 g (1 cos i ) where (1 cos i ) is the distance above the lowest point. Just before the collision, the en ergy of sphere 1 is all kinetic energy, 1 2 m 1 v 2 1 i . Equating the two energies gives 1 2 m 1 v 2 1 i = m 1 g (1 cos i ) . Solving for V 1 i gives v 1 i = { 2 g [1 cos i ] } 1 / 2 = braceleftbig 2 (9 . 8 m / s 2 ) (10 . 2 m) [1 cos(49 )] } 1 / 2 = 8 . 29221 m / s . Question 3, chap 10, sect 99. part 2 of 3 8 points After lighter sphere is let go and collides with the heavier sphere at the bottom of its swing, two spheres immediately bind to gether. What is the speed of the combined system just after the collision? Correct answer: 3 . 63974 m / s (tolerance 1 %). Explanation: This is a perfectly inelastic collision. The speed of the two spheres after collision is de termined by momentum conservation m 1 v 1 i = ( m 1 + m 2 ) V f (1) where m 1 is the mass and v 1 i is the initial velocity of sphere 1 just before the collision, m 2 is the mass of sphere 2, and V f is the velocity of the combined spheres just after...
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 Spring '08
 Turner
 Friction, Mass

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